`x-4y+3z-2w=9`

`3x-2y+z-4w=-13`

`-4x+3y-2z+w=-4`

`-2x+y-4z+3w=-10`

The above system of equations can be represented by the coefficient matrix A and right hand side matrix b as follows:

A=`[[1,-4,3,-2],[3,-2,1,-4],[-4,3,-2,1],[-2,1,-4,3]]`

b=`[[9],[-13],[-4],[-10]]`

The augmented matrix can be written as,

`[[A,b]]=[[1,-4,3,-2,9],[3,-2,1,-4,-13],[-4,3,-2,1,-4],[-2,1,-4,3,-10]]`

Now lets, perform the various row operations to bring the above matrix in the row-echelon...

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`x-4y+3z-2w=9`

`3x-2y+z-4w=-13`

`-4x+3y-2z+w=-4`

`-2x+y-4z+3w=-10`

The above system of equations can be represented by the coefficient matrix A and right hand side matrix b as follows:

A=`[[1,-4,3,-2],[3,-2,1,-4],[-4,3,-2,1],[-2,1,-4,3]]`

b=`[[9],[-13],[-4],[-10]]`

The augmented matrix can be written as,

`[[A,b]]=[[1,-4,3,-2,9],[3,-2,1,-4,-13],[-4,3,-2,1,-4],[-2,1,-4,3,-10]]`

Now lets, perform the various row operations to bring the above matrix in the row-echelon form,

Rewrite the 2nd Row `(R_2)` as `(R_2-3R_1)`

`[[1,-4,3,-2,9],[0,10,-8,2,-40],[-4,3,-2,1,-4],[-2,1,-4,3,-10]]`

Rewrite the 3rd Row`(R_3)` as`(R_3+4R_1)`

`[[1,-4,3,-2,9],[0,10,-8,2,-40],[0,-13,10,-7,32],[-2,1,-4,3,-10]]`

Rewrite the 4th Row`(R_4)` as`(R_4+2R_1)`

`[[1,-4,3,-2,9],[0,10,-8,2,-40],[0,-13,10,-7,32],[0,-7,2,-1,8]]`

Rewrite the 2nd Row`(R_2)` as`(2(R_2+R_3)-R_4)`

`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,-13,10,-7,32],[0,-7,2,-1,8]]`

Rewrite the 3rd Row`(R_3)` as`(R_3+13R_2)`

`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,36,-124,-280],[0,-7,2,-1,8]]`

Rewrite the 4th Row`(R_4)` as `(R_4+7R_2)`

`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,36,-124,-280],[0,0,16,-64,-160]]`

Rewrite the 3rd Row`(R_3)` as `(R_3-R_4)`

`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,20,-60,-120],[0,0,16,-64,-160]]`

Rewrite the 3rd Row by dividing it with 20,

`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,16,-64,-160]]`

Rewrite the 4th Row by dividing it with 16,

`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,1,-4,-10]]`

Rewrite the 4th Row as `(R_3-R_4)`

`[[1,-4,3,-2,9],[0,1,2,-9,-24],[0,0,1,-3,-6],[0,0,0,1,4]]`

Now the matrix is in row-echelon form, and we can perform the back substitution on the corresponding system,

`x-4y+3z-2w=9` ----- Eq:1

`y+2z-9w=-24` ----- Eq:2

`z-3w=-6` ----- Eq:3

`w=4`

Substitute back the value of w in Eq:3,

`z-3(4)=-6`

`z-12=-6`

`z=-6+12`

`z=6`

Substitute back the value of w and z in Eq:2,

`y+2(6)-9(4)=-24`

`y+12-36=-24`

`y=-24+36-12`

`y=0`

Substitute back the value of w,z and y in Eq:1,

`x-4y+3z-2w=9`

`x-4(0)+3(6)-2(4)=9`

`x+18-8=9`

`x=9+8-18`

`x=-1`

So the solutions are x=-1,y=0,z=6 and w=4